bài tích phân này làm sao đậy mấy bạn

D

dien0709

$ I=\int_{0}^{\dfrac{\pi}{6}}\dfrac{tan(x+\dfrac{\pi}{4}}{cos2x}dx$

$ tan(x+\pi/4)=\dfrac{sin(x+\pi/4)}{cos(x+\pi/4)}=\dfrac{sinx+cosx}{cosx-sinx}$

=> $ I=\int_{0}^{\pi/6}\dfrac{dx}{(cosx-sinx)^2}=\int_{0}^{\pi/6}\dfrac{2d(x+\pi/4)}{cos^2(x+\pi/4)}$

=>$ I=2tan(x+\pi/4)|_0^{\pi/6}$
 
You must log in or register to reply here.
Top Bottom